Question

Encontre a integral indefinida das seguintes questões:

Answer 1

$\int\limits {7x^{( \frac{5}{2} )}} +4\, dx = \frac{7x^{ (\frac{5}{2} +1)}}{\frac{5}{2} +1} +4x\\\\ \frac{7x^\frac{7}{2} }{ \frac{7}{2} +4x } = \frac{2*7x^\frac{7}{2} }{7} +4x\\\\=2x^\frac{7}{2} +4=\boxed{2 \sqrt[2]{x^7} +4}\to 2( \sqrt{x^7} +2)+C$

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$\int\limits { \frac{x+1}{x^5} } \, dx = \boxed{\int\limits { \frac{x}{x^5} } \, dx + \int\limits { \frac{1}{x^5} } \, dx }$

resolvendo a primeira 
$\int\limits { \frac{x}{x^5} } \, dx = \int\limits {x^{-4}} \, dx =| \frac{x^{-3}}{-3} |= \frac{1}{-3x^3} =\boxed{- \frac{1}{3x^3} }$

resolvendo a segunda integral
$\int\limits { \frac{1}{x^5} } \, dx = \int\limits {x^{-5}} \, dx =| \frac{x^{-4}}{-4} |=\boxed{- \frac{1}{4x^4} }$

somando as duas
$-\frac{1}{3x^3} - \frac{1}{4x^4} = \frac{-4x-3}{12x^4} = \boxed{\frac{-(4x+3)}{12x^4}+C }$

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$\frac{t^5}{2} - \frac{4}{t^{-3}} +3t = \frac{t^5}{2} -4t^3+3t= \frac{t^5-8t^3+6t}{2} =\boxed{ \frac{1}{2} (t^5-8t^3+6t)}$

agora montando a integral
$\int\limits {\frac{1}{2} (t^5-8t^3+6t)} \, dt = \frac{1}{2} \int\limits {(t^5-8t^3+6t)} \, dt$

integrando

$(t^5-8t^3+6t) = \frac{t^6}{6} -2t^4+3t= \frac{t^6-12t^4+18t}{6} \\\\\\ \frac{1}{2} *\frac{t^6-12t^4+18t}{6} = \boxed{\frac{t^6-12t^4+18t}{12}+C }$

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$(-2+v^{-2})^2\\\\= (-2+ \frac{1}{v^2})^2\\\\(-2)^2+2*(-2)*( \frac{1}{v^2} )+( \frac{1}{v^2} )^2\\\\4- \frac{4}{v^2} + \frac{1}{v^4} = \boxed{4-4v^{-2}+v^{-4}}$

agora integrando

$\int\limits {4-4v^{-2}+v^{-4}} \, dv \\\\=|4v- \frac{4v^{-1}}{-1} + \frac{v^{-3}}{-3} |\\\\=\boxed{4v+ \frac{4}{v}- \frac{1}{3v^3} +C}$

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$n^3(-2n+n^{-5})=-2^4+n^{-2}$


$\int\limits {-2n^4+n^{-2}} \, dn\\\\\ =|-2 \frac{n^5}{5} + \frac{n^{-1}}{-1} |\\\\= \frac{-2n^5}{5} - \frac{1}{n}\\\\ \frac{-2n^{6}-5}{5n} =\boxed{ \frac{-(2n^6+5)}{5n}+C }$

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$\frac{1}{s^4} =s^{-4}\\\\\\ \int\limits{s^{-4}} \, ds = \frac{s^{-3}}{-3} =\boxed{- \frac{1}{3s^3}+C }$

Answer 2

Resposta:

Explicação passo-a-passo: